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2 years ago

Find a solution to the equation 4(x + 7) = 38

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

6 0

Answer:

4(x+7)=38

4x+28=38

4x=38-28

4x=10

(divide by 4)

x=2whole number 2/4

=2whole number 1/2

or x=2.5

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Identify the zero that has a multiplicity greater than one:<br> f(x) = (x - 2)(x + 2)(x – 5)

nadya68 [22]

Answer:

2(multiplicity 2)

Step-by-step explanation:

2 is repeated twice in the function.

4 0

3 years ago

Please help with 11c!?

Contact [7]

Answer:

"To the nearest year, it would be about 9 years"

Step-by-step explanation:

11c)

This is compound growth problem. It goes by the formula:

$F=P(1+r)^t$

Where

F is the future amount

P is the present (initial) amount

r is the rate of growth, in decimal

t is the time in years

Given,

P = 20,000

r = 8% = 8/100 = 0.08

F = double of initial amount = 2 * 20,000 = 40,000

We need to find t:

$F=P(1+r)^t\\40,000=20,000(1+0.08)^t\\2=(1.08)^t$

To solve exponentials, we can take Natural Log (Ln) of both sides:

$2=(1.08)^t\\Ln(2)=Ln((1.08)^t)$

Using the rule shown below we can simplify and solve:

$Ln(a^b)=bLn(a)$

We can write:

$Ln(2)=Ln((1.08)^t)\\Ln(2)=tLn(1.08)\\t=\frac{Ln(2)}{Ln(1.08)}\\t=9.0064$

To the nearest year, that would be about 9 years

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3 years ago

The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who w

dimulka [17.4K]

Answer:

99% CI:

$-4.637\leq\mu_v-\mu_o\leq 3.737$

Step-by-step explanation:

We have to calculate a 99%CI for the difference of means for the vegan and the omnivore.

First, we have to estimate the standard deviation

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}$

The MSE can be calculated as

$MSE=\frac{(SSE_1+SSE_2)}{df} =\frac{(n_1s_1)^2+(n_2*s_2)^2}{n_1+n_2-2}\\\\MSE=\frac{(85*1.05)^2+(96*1.20)^2}{85+96-2}=\frac{7965.56+13272.04}{179} =118.65$

The weighted sample size nh can be calculated as

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{1/85+1/96}=\frac{2}{0.0222} =90.16$

The standard deviation then becomes

$s_{Md}=\sqrt{\frac{2MSE}{n_h}}=\sqrt{\frac{2*118.65}{90.16}}=\sqrt{2.632} =1.623$

The z-value for a 99% confidence interval is z=2.58.

The difference between means is

$\Delta M=M_v-M_o=5.10-5.55=-0.45$

Then the confidence interval can be constructed as

$\Delta M-z*s_{Md}\leq\mu_v-\mu_o\leq \Delta M+z*s_{Md}\\\\ -0.45-2.58*1.623\leq\mu_v-\mu_o\leq -0.450+2.58*1.623\\\\-0.450-4.187\leq\mu_v-\mu_o\leq-0.450+4.187\\\\-4.637\leq\mu_v-\mu_o\leq 3.737$